Let’s break this down step by step to construct the requested functions based on $h(t) = e^t$.

(a) First shifted five units right.

Shifting the graph five units to the right means replacing $t$ with $t - 5$.

Thus: $h_a(t) = e^{t - 5}.$

(b) Then reflected across the $t$-axis.

Reflection across the $t$-axis negates the entire function $h(t)$. That means multiplying $e^t$ by $-1$.

Thus: $h_b(t) = -e^{t}.$

(c) Then shifted down by seven units.

To shift a function down by seven units, subtract 7 from the function. Applying this to $h_b(t) = -e^t$:

$h_c(t) = -e^t - 7.$

Final Answers:

(a) $h_a(t) = e^{t - 5}$
(b) $h_b(t) = -e^t$
(c) $h_c(t) = -e^t - 7$

Would you like further clarification on any step, or do you need a graph to visualize these transformations?

Here are 5 related questions you might find useful:

1. How does shifting left instead of right affect the function?
2. What happens to the graph if we reflect it across the $y$-axis instead?
3. Can you explain why subtracting 7 shifts the graph downward?
4. How would combining multiple transformations in a different order change the final graph?
5. What is the domain and range of the original and transformed functions?

Tip: When transforming exponential functions, always modify the exponent or function output carefully to preserve the base's meaning!